# What stage of growth are the cells I counted? - (Nov/13/2013 )

Hi everyone, I'm new to the forum. I'm first year biochemistry at the University of Bristol. Using a haemocytometer I counted cell no. of three different cell cultures (colorectal cancer), one negative control, one + known effective anticancer drug, and one + new anticancer drug.

This question relates to this is:

The colorectal cancer cells (HT29) that you counted in experiment 1 were in the lag/log/stationary/death stage of the growth curve.

Can anyone tell me how I work this out? Should I plot a growth curve? Any help is appreciated.

Thanks.

Do you have multiple cell counts at different time points or are you just given one number? Go to ATCC and find the growth curve of your cells and from here you can determine what phase your cells were in.

__ATCC__:

http://physics.cancer.gov/docs/bioresource/colorectal/NCI-PBCF-HTB38_HT-29_SOP-508.pdf

jerryshelly1 on Wed Nov 13 15:00:01 2013 said:

Do you have multiple cell counts at different time points or are you just given one number? Go to ATCC and find the growth curve of your cells and from here you can determine what phase your cells were in.

ATCC:http://physics.cancer.gov/docs/bioresource/colorectal/NCI-PBCF-HTB38_HT-29_SOP-508.pdf

Cool, thanks mate. Yes, only one cell count.

jerryshelly1 on Wed Nov 13 15:00:01 2013 said:

Do you have multiple cell counts at different time points or are you just given one number? Go to ATCC and find the growth curve of your cells and from here you can determine what phase your cells were in.

ATCC:http://physics.cancer.gov/docs/bioresource/colorectal/NCI-PBCF-HTB38_HT-29_SOP-508.pdf

I found the growth profile on the link you posted but am not sure how to translate that into lag/log/stationary/death? My data says that the negative control had a cell count of 3.75x10^6, known drug 2.80x10^6 and new drug 1.05x10^6.

Just realised that the graph I plotted from a list of data points was in relation to my question, so the answer was the cells are in log phase of growth. Thanks for your help.